Question

The block shown in figure is released from rest when the extension in spring (of spring constant $k$) is $2\text{m}$. Find out the speed of the block when the spring is compressed by $1\text{m}$.

• A. $\sqrt{\frac{k}{2}}$
  
• B. $\sqrt{\frac{3k}{2}}$
  
• C. $\sqrt{\frac{5k}{2}}$
  
• D. $\sqrt{\frac{7k}{2}}$
  

Answer 1

The correct option is B $\sqrt{\frac{3k}{2}}$



In the above situation, the only force acting on the block is spring force which is conservative in nature.

So, applying mechanical energy conservation between initial and final positions we have:

$k_f+u_f=k_i+u_i$ .............$\left(1\right)$

Initially, the block is at rest, $k_i=0$
$u_i=\frac{1}{2}k{x}_1^2=\frac{1}{2}k(2{)}^2=2k$ $\text{Joule}$

Finally, at position $\text{B}$, if the speed of the block is $v$ then
$k_f=\frac{1}{2}m{v}^2=\frac{1}{2}\times 2\times v^2=v^2$ $\text{Joule}$

$u_f=\frac{1}{2}k(x_2{)}^2=\frac{1}{2}\times k\times 1=\frac{k}{2}$ $\text{Joule}$

Putting the above values in equation $\left(1\right)$, we get

$\Rightarrow v^2+\frac{k}{2}=2k\Rightarrow v^2=\frac{3k}{2}\Rightarrow v=\sqrt{\frac{3k}{2}}\text{m/sec}$

Why this Question ? Key point :Conservation of mechanical energy can only be applied when only conservative forces are acting. If non-conservative forces are acting on a system, its always advisable to use work-kinetic energy theorem, which is given by, $W_{net}=\Delta K.E$